What is the integral from 0 to 1 of (2-2x)^2 ?

The integral from 0 to 1 of (2-2x)^2 is 4/3

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integral from 0 to 1 of (2-2x)^2

\int_{0}^{1} (2 - 2 x)^{2} d x

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

\int_{0}^{1} (2 - 2 x)^{2} d x

Apply u-substitution

: \int_{2}^{0} - \frac{u ^{2}}{2} d u

= \int_{2}^{0} - \frac{u ^{2}}{2} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{2} - \frac{u ^{2}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{2} \cdot \int_{0}^{2} u^{2} d u)

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{2}

= - (- \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{2})

Simplify

= \frac{1}{2} [\frac{u ^{3}}{3}]_{0}^{2}

Compute the boundaries:

\frac{8}{3}

= \frac{1}{2} \cdot \frac{8}{3}

Simplify

= \frac{4}{3}

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What is the integral from 0 to 1 of (2-2x)^2 ?
The integral from 0 to 1 of (2-2x)^2 is 4/3